HCID 131725237, compound → HCID 10967292, compound
Reaction equation
[NH:1]([N:18]=[N+:19]=[N-])[C@H:2]([C:8]([NH:10][CH2:11][C:12]1[CH:17]=[CH:16][CH:15]=[CH:14][CH:13]=1)=[O:9])[CH2:3][C:4](=[O:7])[O:5][CH3:6].N(N=[N+]=[N-])[C@H](C(O)=O)CC(=O)OC>>[CH3:6][O:5][C:4](=[O:7])[CH2:3][C@H:2]([N:1]=[N+:18]=[N-:19])[C:8]([NH:10][CH2:11][C:12]1[CH:17]=[CH:16][CH:15]=[CH:14][CH:13]=1)=[O:9]Participants
Procedure
Benzyl-protected L-aspartate was used in the procedure of Lundquist and Pelletier17 to give N3(L)Asp(OMe)OH (46-L). Under these conditions, we observed transesterification to give the methyl ester product as opposed to the benzyl ester. N3(L)Asp(OMe)OH (46-L) was produced as a yellowish oil in 78% yield. N3(L)Asp(OMe)OH (46-L) was then coupled with benzylamine as above to give N3(L)Asp(OMe)NHBn (49-L) as a yellowish oil in 90% yield (70% overall, two steps). The procedure above was repeated with benzyl-protected D-aspartate to give N3(L)Asp(OMe)NHBn (49-D) as a yellowish oil in 67% overall yield. Spectral Data. The spectral data for both N3Asp(OMe)OH (D and L enantiomers) and both N3Asp(OMe)NHBn (D and L enantiomers) products are identical. N3(L)Asp(OMe)OH (46-L) 1H NMR (500 MHz, CDCl3) δ 10.24 (bs,1H), 4.47 (dd, J=7.4, 5.3 Hz, 1H), 3.76 (s, 3H), 2.91 (dd, J=16.9, 5.1 Hz, 1H), 2.79 (dd, J=16.8, 7.6 Hz) ppm; 13C NMR (125 MHz, CDCl3) δ 174.68, 170.12, 50.09, 52.44, 35.84 ppm; MS (ESI) m/z 196.0340 (MNa+[C5H7N3O4Na]=196.0334). N3(L)Asp(OMe)NHBn (49-L) 1H NMR (500 MHz, CDCl3) δ 7.38-7.27 (m, 5H), 6.83 (bs, 1H), 4.54 (m, 3H), 3.75 (s, 3H), 3.18 (dd, J=17.1, 3.7 Hz, 1H), 2.75 (dd, J=17.3, 8.7 Hz, 1H) ppm; 13C NMR (125 MHz, CDCl3) δ 170.77, 167.90, 137.35, 128.81, 127.80, 127.77, 60.32, 52.24, 43.71, 37.00 ppm; MS (ESI) m/z 285.0953 (MNa+[C12H12N4O3Na]=285.0964).
Sources & evidence
- ORD record
- ord-baea46a7deb041a591411e5b2cf4eddf
- Source dataset
- uspto-grants-2008_01
- Patent number
- US07317129B2
Reaction notes
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